Problem: Find the slope and y-intercept of the line that is ${\text{parallel}}$ to $\enspace {y = -\dfrac{1}{3}x + 2}\enspace$ and passes through the point ${(2, -7)}$. {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9} {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9}
Explanation: Parallel lines have the same slope. The slope of the blue line is ${-\dfrac{1}{3}}$ , so the equation of our parallel line will be of the form $\enspace {y = -\dfrac{1}{3}x + b}\enspace$ We can plug our point, $(2, -7)$ , into this equation to solve for ${b}$ , the y-intercept. $-7 = {-\dfrac{1}{3}}(2) + {b}$ $-7 = -\dfrac{2}{3} + {b}$ $-7 + \dfrac{2}{3} = {b} = -\dfrac{19}{3}$ The equation of the parallel line is $\enspace {y = -\dfrac{1}{3}x - \dfrac{19}{3}}\enspace$. ${m = -\dfrac{1}{3}, \enspace b = -\dfrac{19}{3}}$